The displacement x in meter of a particle of mass m kg moving in one direction under the action of a force is related to the time t in second by the equation :
\(x = (t - 3)^2\)
The work done by the force (in joules) in first six seconds is :
Answer & explanation
Correct answer: option 4
According to ques : \(x = (t-3)^2\)
\(x = t^2 - 6t + 9\)
\(v = \frac{dx}{dt} = 2t - 6\)
At t = 0, v = 2 x 0 - 6 = -6
at t = 6 s, v = 2 x 6 - 6 = 6
Initial and final KE are same,
From Work Energy Theorem , Work Done is equal to change in K.E.
$W = \Delta K$
hence no work is done.